Question

Find the equation of the parabola whose axis is parallel to $$y$$-axis which passes through points $$(4 , 5)$$,$$(-2 , 11)$$ and $$(-4 , 21)$$.

Solution

A parabola with axis parallel to y-axis has equal to $$y=ax^2+bx+c$$, for constants a, b, c in each of the three points.$$5=a(4)^2+b(4)+c$$$$11=a(-2)^2+b(-2)+c$$$$21=a(-4)^2+b(-4)+c$$So, $$\left(\begin{matrix} 16 & 4 & 1 \\ 4 & -2 & 1\\ 16 & -4 & 1\end{matrix}\right|\left.\begin{matrix} 5 \\ 11\\ 21 \end{matrix}\right)$$ is our system.This simplifies as follows$$\Rightarrow \left(\begin{matrix} 16 & 4 & 1\\ 4 & -2 & 1\\ 0 & -8 & 0\end{matrix}\right|\left.\begin{matrix} 5 \\ 11\\ 16\end{matrix}\right)$$$$\Rightarrow \left(\begin{matrix} 1 & 1/4 & 1/16 \\ 0 & -3 & 3/4\\ 0 & 1 & 0\end{matrix}\right|\left.\begin{matrix} 5/16\\ 39/4\\ -2\end{matrix}\right)$$$$\Rightarrow \left(\begin{matrix} 1 & 0 & 1/16\\ 0 & 1 & 0\\ 0 & 0 & 1\end{matrix}\right|\left.\begin{matrix} 13/16\\ -2 \\ 5\end{matrix}\right)$$Now we get$$c=5$$$$b=2$$$$a=\dfrac {13}{16}-5\dfrac{1}{16}$$$$=\dfrac{13-5}{16}$$$$=\dfrac{8}{16}$$$$=\dfrac{1}{2}$$Finally we write $$y=\dfrac{1}{2}.x^2-2x+5$$.Maths

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